By definition we have h=u+pv for a system where h=specific enthalpy u=specific internal energy p=external pressure applied on the system v=specific volume of the system The internal energy is arbitrarily taken to be zero at the triple point of water i.e, u=0 when T=273.16K and P=4.587 mm of Hg. So at the triple point of water we have enthalpy of water to be slightly positive due to small value of the 'pv' term. At freezing point,however,we have T=273.16K and P=760 mm of Hg. So the change in enthalpy could be obtained as Δh=Δu+Δ(pv) Now, Δu=0 as ΔT=0 and the only difference in enthalpy is due to the Δ(pv) . At triple point, we have a range of values for 'v' along the triple point line and p=4.587 mm of Hg. At freezing point, we have v=0.001 m3/kg and p=760 mm of Hg. So to bring a unit mass of water from triple point to the freezing point a reversible isothermal path could be followed at T=273.16K from P1=4.587 mm of Hg to P2=760 mm of Hg. Thus the change in enthalpy would be equal to the displacement work done to bring water from the triple point to the freezing point i.e, Δ(pv). But in most text books the enthalpy of water at both triple and freezing points are taken to be equal.Whereas they cannot be theoretically equal in value is this because the change in specific volume is too small to take Δ(pv) into account?